Javascript: SyntaxError: await is only valid in async function

Question

I am on Node 8 with Sequelize.js

Gtting the following error when trying to use await.

SyntaxError: await is only valid in async function

Code:

async function addEvent(req, callback) {
    var db = req.app.get('db');
    var event = req.body.event

    db.App.findOne({
        where: {
            owner_id: req.user_id,
        }
    }).then((app) => {

                let promise = new Promise((resolve, reject) => {
                    setTimeout(() => resolve("done!"), 6000)

                })

               // I get an error at this point 
               let result = await promise;

               // let result = await promise;
               //              ^^^^^
               // SyntaxError: await is only valid in async function
            }
    })
}

Getting the following error:

               let result = await promise;
                            ^^^^^
               SyntaxError: await is only valid in async function

What am I doing wrong?

then((app) => { this anonymous function is not marked as async — skyboyer
– skyboyer, Commented Oct 24, 2018 at 17:49
Probably better to just return the promise in the then callback and move it down the chain rather than awaiting it... — Clint
– Clint, Commented Oct 24, 2018 at 17:50
I find it kinda wierd the way your trying to get the promise resolve, why your not using .then() ? — Talg123
– Talg123, Commented Oct 24, 2018 at 17:50

Ankush Sharma · Accepted Answer · 2018-10-24 17:49:29Z

18

You can run await statement only under async function. https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Statements/async_function

So, you can write your

}).then((app) => {

as

}).then(async (app) => {

answered Oct 24, 2018 at 17:49

Ankush Sharma

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2 Comments

user1107173 Over a year ago

Thank you for your answer! You were the first, but b/c of clarity and depth in @estus answer, I marked it as correct!

Zoette Over a year ago

Yaaaaay you made my day :D :D :D

Estus Flask · Accepted Answer · 2018-10-24 18:01:31Z

addEvent is a mixture of async..await and raw promises. await is syntactic sugar for then. It's either one or another. A mixture results in incorrect control flow; db.App.findOne(...).then(...) promise is not chained or returned and thus is not available from outside addEvent.

It should be:

async function addEvent(req, callback) {
    var db = req.app.get('db');
    var event = req.body.event

    const app = await db.App.findOne({
        where: {
            owner_id: req.user_id,
        }
    });

    let promise = new Promise((resolve, reject) => {
        setTimeout(() => resolve("done!"), 6000)
    })

    let result = await promise;
}

Generally plain callbacks shouldn't be mixed with promises. callback parameter indicates that API that uses addEvent may need to be promisified as well.

user5047085 · Accepted Answer · 2018-10-24 17:49:05Z

3

async/await only works if the immediate function has the async keyword, you need this:

...
    }).then(async app => {   // <<<< here

                let promise = new Promise((resolve, reject) => {
                    setTimeout(() => resolve("done!"), 6000)

                })

               // I get an error at this point 
               let result = await promise;

               // let result = await promise;
               //              ^^^^^
               // SyntaxError: await is only valid in async function
            }
    })

answered Oct 24, 2018 at 17:49

user5047085

Comments

Tibin Thomas · Accepted Answer · 2018-10-24 17:59:12Z

1

You can use await only inside a function which is async. Also you can await only a piece of code that returns a promise.

Here you are using await inside a different context. Better you use then() here to solve the problem.

await only works if the immediate function that encloses it is async.

edited Oct 24, 2018 at 17:59

answered Oct 24, 2018 at 17:51

Tibin Thomas

7826 silver badges12 bronze badges

1 Comment

Clint Over a year ago

How does this answer the question? You basically just restated what the error message says.

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